Problem 43 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

LABOR EFFICIENCY A company wishes to examine the efficiency of two members of its senior staff, Jack and Jill, who work independently of one another. Let \(X\) and \(Y\) be random variables that measure the proportion of the work week that Jack and Jill, respectively, actually spend performing their duties. Assume that the joint probability density function for \(X\) and \(Y\) is \(f(x, y)= \begin{cases}0.4(2 x+3 y) & \text { if } 0 \leq x \leq 1,0 \leq y \leq 1 \\ 0 & \text { otherwise }\end{cases}\) a. Verify that \(f(x, y)\) satisfies the requirements for a joint probability density function. b. Find the probability that Jack spends less than half his time working while Jill spends more than half her time working. c. Find the probability that Jack and Jill each spend at least \(80 \%\) of the work week performing their assigned tasks. d. Find the probability that Jack and Jill combine for less than a full work week. [Hint: This is the event that \(X+Y<1\).]

Step-by-Step Solution

Answer

Part a: Verified.Part b: 0.275.Part c: 0.072.Part d: 0.4667.

Step 1: Verify Joint Probability Density Function

To verify that the given function is a joint probability density function, integrate it over the entire range of possible values for X and Y. We need to check that the integral of the joint pdf over its entire range equals 1: \[ \int_{0}^{1} \int_{0}^{1} 0.4(2x + 3y) \, dx \, dy \]

Step 2: Solve Inside Integral

Integrate the function with respect to x first: \[ \int_{0}^{1} 0.4(2x + 3y) \, dx = 0.4 \left[ x^2 + 3yx \right]_0^1 = 0.4(1 + 3y) \]

Step 3: Solve Outside Integral

Now, integrate the resulting function with respect to y: \[ \int_{0}^{1} 0.4(1 + 3y) \, dy = 0.4 \left[ y + \frac{3y^2}{2} \right]_0^1 = 0.4 \left( 1 + \frac{3}{2} \right) = 1 \] Therefore, the joint probability density function is valid.

Step 4: Probability Calculation for Part b

To find the probability that Jack spends less than half his time working while Jill spends more than half her time working, evaluate: \[ P(X < 0.5, Y > 0.5) = \int_{0}^{0.5} \int_{0.5}^{1} 0.4(2x + 3y) \, dy \, dx \]

Step 5: Solve Inner Integral for Part b

Integrate the inner part with respect to y first: \[ \int_{0.5}^{1} 0.4(2x + 3y) \, dy = 0.4 \left[ 2xy + \frac{3y^2}{2} \right]_{0.5}^{1} = 0.4 (2x + \frac{3}{2} - x - \frac{3}{8}) = 0.4 (x + \frac{9}{8}) \]

Step 6: Solve Outer Integral for Part b

Next, integrate the result with respect to x: \[ \int_{0}^{0.5} 0.4 \left( x + \frac{9}{8} \right) \, dx = 0.4 \left[ \frac{x^2}{2} + \frac{9x}{8} \right]_{0}^{0.5} = 0.4 \left( \frac{1}{8} + \frac{9}{16} \right) = 0.4 \left( \frac{2+9}{16} \right) = 0.4(\frac{11}{16}) = 0.275 \]

Step 7: Probability Calculation for Part c

To find the probability that Jack and Jill each spend at least 80% of the work week performing their assigned tasks: \[ P(X \geq 0.8, Y \geq 0.8) = \int_{0.8}^{1} \int_{0.8}^{1} 0.4(2x + 3y) \, dx \, dy \]

Step 8: Solve Inner Integral for Part c

Integrate the inner part with respect to x first: \[ \int_{0.8}^{1} 0.4(2x + 3y) \, dx = 0.4 \left[ x^2 + 3yx \right]_{0.8}^{1} = 0.4((1 + 3y) - (0.64 + 2.4y)) = 0.4(0.36 + 0.6y) \]

Step 9: Solve Outer Integral for Part c

Next, integrate the result with respect to y: \[ \int_{0.8}^{1} 0.4(0.36 + 0.6y) \, dy = 0.4 \left[ 0.36y + 0.3y^2 \right]_{0.8}^{1} = 0.4 \left( 0.36 + 0.3 - 0.288 - 0.192 \right) = 0.4 (0.18) = 0.072 \]

Step 10: Probability Calculation for Part d

To find the probability that Jack and Jill combine for less than a full work week, we need to evaluate: \[ P(X + Y < 1) = \int \int_{x + y < 1} 0.4(2x + 3y) \, dx \, dy \] This geometry forms a triangular region.

Step 11: Determine Region Boundaries

The boundaries for the region where \[ x + y < 1 \] are: 0 <= x <= 1 0 <= y <= 1-x

Step 12: Calculate Integral for Region

To find the probability, calculate the double integral over the defined region: \[ \int_{0}^{1} \int_{0}^{1-x} 0.4 (2x + 3y) \, dy \, dx \] Begin with the inner integral for dy: \[ \int_{0}^{1-x} 0.4 (2x + 3y) \, dy = 0.4 \left[ 2xy + \frac{3y^2}{2} \right]_{0}^{1-x} = 0.4 \left(2x(1-x) + \frac{3(1-x)^2}{2}\right) \] Simplify: \[ = 0.4 \left(2x - 2x^2 + \frac{3(1 - 2x + x^2)}{2}\right) = 0.4 \left(2x - 2x^2 + \frac32 - 3x + \frac{3x^2}{2}\right) \] \[= 0.4 \left(\frac{3}{2} - x - 0.5x^2\right) \]

Step 13: Solve Outer Integral

Next, integrate with respect to x: \[ \int_{0}^{1} 0.4 \left(\frac{3}{2} - x - 0.5x^2\right) \, dx = 0.4 \left[ \frac{3x}{2} - \frac{x^2}{2} - \frac{x^3}{6} \right]_{0}^{1} = 0.4 \left(1.5 - 0.5 - \frac{1}{6}\right) = 0.4 \left(\frac{2.5 - 0.1667}{1}\right) = 0.4667 \]

Key Concepts

Random Variables

In this problem, we are dealing with two random variables, Jack and Jill's efficiency, denoted as X and Y, respectively. A random variable is a variable whose possible values are numerical outcomes of a random phenomenon. In simpler terms, it's a way to quantify random events. Here, X represents the proportion of the work week Jack spends working, and Y represents the same for Jill. These values can range between 0 and 1.

Integration

Integration is a fundamental concept in calculus used to find areas under curves, among other things. In the context of probability, it helps compute probabilities over a range of values. The integration steps in the solution include calculating the joint probability density function (pdf), integrating first with respect to x and then with respect to y. Remember, the double integral over the entire range of our joint pdf must equal 1 to be valid:
\[ \int_{0}^{1} \int_{0}^{1} 0.4(2x + 3y) \, dx \, dy = 1 \]] The integration steps involve solving one variable at a time, simplifying the inner integral first, and then tackling the outer integral.

Probability Calculation

Probability calculation in this context involves integrating the joint pdf over a specified range. For example, for **Finding Jack's and Jill's specific working hours:**
\[ P(X < 0.5, Y > 0.5) = \int_{0}^{0.5} \int_{0.5}^{1} 0.4(2x + 3y) \, dy \, dx \]] This means we calculate the probability that Jack spends less than half of his work week working while Jill spends more than half. The steps for these calculations typically involve:

• Defining the integral's limits to match the problem.
• Integrating with respect to one variable.
• Integrating the resulting expression with respect to the second variable.

The goal is to compute the area under the probability density function within the given limits, which represents the desired probability.

Joint PDF Validation

For a function to be a joint probability density function (pdf), it needs to meet several criteria:

• The function must be non-negative for all values within its range.
• The integral of the joint pdf over its entire range must equal 1.

In our example, we verified that:
\[ \int_{0}^{1} \int_{0}^{1} 0.4(2x + 3y) \, dx \, dy = 1 \]] This ensures the given function is valid as it sums up to 1 when integrated over all possible values of X and Y, confirming it's a valid joint pdf. This criterion ensures that we're actually dealing with probabilities that make sense and obey the fundamental rules of probability theory.